7
$\begingroup$

Question

So recently I was thinking about this: How many scalars are available in $4$ dimensions in General Relativity (without being redundant)? For example, with metric we can construct the following scalar:

$$ g^{\mu \nu} g_{\mu \nu} = 4 $$ is the same as: $$ (g^{\mu \nu} \otimes g^{\rho \kappa}) \cdot (g_{\mu \nu} \otimes g_{\rho \kappa} ) = 16 $$

We also have scalars like curvature, torsion, inner product of the riemann tensor with itself, etc.

Motivation

My motivation for doing so is as follows: GR is currently through (rank $2$ symmetric) tensors formulated as: $$ R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} $$ Hence any solution of the above automatically satisfies:
$$ (R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu}) (R^{\mu \nu} - \frac{1}{2} R g^{\mu \nu}) = \bigg(\frac{8 \pi G}{c^4}\bigg)^2 T_{\mu \nu} T^{\mu \nu} $$

But note the later equation is written purely in invariant observables. I was wondering if General Relativity could also be written purely in terms of observables? If not how many short are we? Can the remaining variables be expressed as something invariant and not a scalar (not sure if it would be a tensor either)?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ relevant: Lovelock's theorem. $\endgroup$ – AccidentalFourierTransform Oct 29 '18 at 19:46
  • $\begingroup$ @AccidentalFourierTransform If the reformulation was successful wouldn't the equations be the same just the fundamental variables be different? $\endgroup$ – More Anonymous Oct 29 '18 at 20:03
  • $\begingroup$ There are many cases in which it is useful to determine how quantities change with reference frame, because those quantities are experimentally relevant. Which specific parts of GR are you proposing to recast? $\endgroup$ – probably_someone Oct 29 '18 at 20:57
5
$\begingroup$

If you define invariant observables to be relativistic scalars formed from the Riemann tensor, then the answer is no, it is not possible to formulate GR in terms of invariant observables. As a counterexample, all curvature invariants vanish for any gravitational plane-wave solution in 3+1 dimensions. This is basically because an observer who chases the wave at high velocity can see the energy and amplitude of the wave Doppler shifted down to an arbitrarily low level. For more on this, see Schmidt, http://arxiv.org/pdf/gr-qc/9404037v1.pdf .

Re the classification of scalar invariants formed from the Riemann tensor (without taking derivatives), see https://en.wikipedia.org/wiki/Carminati%E2%80%93McLenaghan_invariants . If you take derivatives, the number of scalar invariants you can form is infinite.

I think what this really tells us is that relativistic scalars are not rich enough to describe all the things we can actually observe in GR. For instance, if a black hole merger results in the radiation of gravitational waves, then it's reasonable to talk about an observable which is the intensity of those waves in the frame of reference of a distant observer at rest with respect to the source. This is an observable that can't be described as a relativistic scalar, but it's clearly an observable -- LIGO did observe such a thing.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ This is an observable that can't be described as a relativistic scalar, but it's clearly an observable. This statement should be clarified, as any observable can be translated into a relativistic scalar if we introduce the observer's four-velocity. $\endgroup$ – Cham Mar 1 '19 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.